Question

Calculate the pH of a solution prepared by adding 25.0 ml of .10 M sodium hydroxide to 30.0 ml of .20 M acetic acid

Answer 1

SOLUTION:-

Since you're dealing with a buffer, which in your case is a solution that contains a weak acid, acetic acid, and its conjugate base, the acetate anion, in comparable amounts, you can use the Henderson-Hasselbalch equation to determine its pH.

PH_sol = pKa+log([conjugate base]/[weak acid])

Here pKa is equal to

pKa=−log(Ka)

pKa=−log(1.76⋅10⁻⁵)=4.75

Use the classic approach to determine the concentrations of the weak acid and of the conjugate base after the sodium hydroxide solution is added.

Since you add together 2.5⋅10⁻³moles of sodium hydroxide and 3⋅10⁻³moles of acetic acid, you will be left with

nacetic acid=3⋅10⁻³−2.5⋅10⁻³=0.5⋅10⁻³moles

The reaction will produce the same number of moles of acetate ions as the number of moles of sodium hydroxide consumed. This means that you will get

[CH3COOH]=3⋅10⁻³moles/55⋅10⁻³L=0.05454 M

[CH3COO−]=2.5⋅ 10⁻³moles/55⋅10⁻³L=0.04545 M

Therefore, the solution's pH is

pHsol=4.75+log(0.04545M/0.05454M)

pHsol=4.75+(−0.0791378)=4.670